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Thread: Imaginary unit "j" in ratio

  1. #1
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    Question Imaginary unit "j" in ratio

    Hi, this is partially a question from physics, but I think mathematicians will also know the answer:

    Why is $\frac{1}{j \omega C} = \frac{-j}{\omega C}$?
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  2. #2
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    Re: Imaginary unit "j" in ratio

    Quote Originally Posted by Nforce View Post
    Hi, this is partially a question from physics, but I think mathematicians will also know the answer:
    Why is $\frac{1}{j \omega C} = \frac{-j}{\omega C}$?
    You may not care for the answer but here it is:
    it is the same reason some insist that $\dfrac{1}{\sqrt 2}$ be written as $\dfrac{\sqrt 2}{2}$.

    Both are holdovers from pre-calculator days when calculations were made easier by rationalizing the denominator. Dividing by two is easy. Same can be said of $\bf{i}$. Moreover $\dfrac{1}{\bf{i}}=\dfrac{-\bf{i}}{1}$ is in fact true.
    Last edited by Plato; Jan 16th 2018 at 06:46 AM.
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  3. #3
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    Re: Imaginary unit "j" in ratio

    Great answer Plato, I completely understand.

    Thanks.
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    Re: Imaginary unit "j" in ratio

    It's counter-intuitive that $\frac1{\mathbb j}=-\mathbb j$, but to see it clearly:
    $$\frac1{\mathbb j} = \frac{\mathbb j}{\mathbb j} \cdot \frac1{\mathbb j} = \frac{\mathbb j}{\mathbb j \cdot \mathbb j} = \frac{\mathbb j}{\mathbb j^2} = \frac{\mathbb j}{-1}=-{\mathbb j}$$
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